Answer:
[tex]sin P=\frac{4}{5}[/tex].
Step-by-step explanation:
In triangle QRP
PQ= 20
PR=12
QR= 16
[tex]sin\theta= \frac{perpendicular\; side }{Hypotenuse}[/tex]
[tex]sinP=\frac{QR}{PQ}[/tex]
[tex]\therefore sinP=\frac{16}{20}[/tex]
[tex]sinP=\frac{4}{5}[/tex]
sinQ=[tex]\frac{PR}{PQ}[/tex]
sinQ=[tex]\frac{12}{20}[/tex]
sinQ=[tex]\frac{ 3}{5}[/tex]
In triangle STU
SU=16
UT=34
ST=30
sinU=[tex]\frac{ST}{TU}[/tex]
sinU=[tex]\frac{30}{34}[/tex]
sinU=[tex]\frac { 15}{17}[/tex]
sinT=[tex]\frac{US}{UT}[/tex]
sinT=[tex]\frac{16}{34}[/tex]
sinT=[tex]\frac {8}{17}[/tex]
Hence, sinP=[tex]\frac{4}{5}[/tex].
